Find the time constant of the following circuit.

$A$ coil of inductance $1\,H$ and resistance $100\,\Omega$ is connected to a battery of $6\,V$. Determine approximately:
$(a)$ The time elapsed before the current acquires half of its steady-state value.
$(b)$ The energy stored in the magnetic field associated with the coil at an instant $15\,ms$ after the circuit is switched on. (Given $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

An inductor of inductance $L = 400 \ mH$ and resistors of resistance $R_1 = 2 \ \Omega$ and $R_2 = 2 \ \Omega$ are connected to a battery of emf $E = 12 \ V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$. The potential drop across $L$ as a function of time is:

An inductor with an iron core is connected in series with a resistor $R$ across an ideal $DC$ source. The circuit is in a steady state. If the iron core is removed, what happens to the current flowing through the inductor immediately after the removal of the core?

In the circuit shown,the switch $S$ is initially open. The switch $S$ is closed at $t = 0$. The difference between the maximum and minimum current that can flow in the circuit is.....$A$.

In the circuit shown below,the switch is closed at time $t=0$. Which of the graphs shown below best represents the voltage across the inductor,as seen on an oscilloscope?